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not sure..

 
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nielskool
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PostPosted: Thu Apr 02, 2009 12:33 pm    Post subject: not sure.. Reply with quote
heey guys i was reading some posts about pointers in c
and found this: http://forum.cheatengine.org/viewtopic.php?t=386636
i don't get it al..
i think i know what it does and how it works(a little)
but i dont get the: &ulValue in the example, what must be placed in the first arg in the function? or does it return an value at that var?
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Flyte
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Location: Canada
PostPosted: Thu Apr 02, 2009 12:50 pm    Post subject: Reply with quote
When you place an ampersand (&) in front of a variable, you can basically read it as: the address of....

Also, when you place an asterisk (*) in front of a variable (outside it's declaration), it means: the value at this address....

Code:
int x = 2;
// Make 'y' point to 'x'. Move the address of 'x' into 'y'.
// Note the asterisk means we are declaring a pointer, not taking
// the value at the address.
int *y = &x;

// Now we use the asterisk to get the value 'y' points at.
printf("y points to: %d\n", *y);
// y points to: 2

// We can also change 'x' by changing the value 'y' points to.
*y = 3;
printf("x is now: %d\n", x);
// x is now: 3


So, the reason we pass a pointer to the variable in the function, is because the function needs to be able to change what is in the variable directly.
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nielskool
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Posts: 65
PostPosted: Thu Apr 02, 2009 12:57 pm    Post subject: Reply with quote
was reading it.. gave you a repp and i thought hmm this code wasn't there yet right..Razz
went back to the page and reloaded it.. you added something.. already got it but thanks, now i am sure

//edit

hmm.. get an error.. what do i have to do with:
Code:

#define ReadPointer(x, y, ...) _ReadPointer(x, y, __VA_ARGS__, OFFSETS_END)

because it gives me:
[Error] C:\Users\niels kool\Documents\C-Free\Temp\Untitled1.cpp:27: badly punctuated parameter list in `#define'
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